orders/index, Prove [G:H]=[G:K][K:H]

apple2009 · #1 $Old$ November 13th, 2009, 08:48 PM

Suppose H and K are both subgroup of a finite group G, and H⊂K. Prove [G:H]=[G:K][K:H]

Jose27 · #2 $Old$ November 13th, 2009, 09:42 PM

Chris L T521 · #3 $Old$ November 13th, 2009, 09:42 PM

Quote:

Originally Posted by apple2009 $View Post$

Suppose H and K are both subgroup of a finite group G, and H⊂K. Prove [G:H]=[G:K][K:H]

In a sense here, we see that $H\leq K\leq G$ . It follows by Lagrange's Theorem that both $\left|H\right|$ and $\left|K\right|$ divide $\left|G\right|$ .

Therefore,

$\left[G:K\right]=\frac{\left|G\right|}{\left|K\right|}\implies \left|G\right|=\left[G:K\right]\left|K\right|$

$\left[G:H\right]=\frac{\left|G\right|}{\left|H\right|}\implies\left|G\right|=\left[G:H\right]\left|H\right|$

$\left[K:H\right]=\frac{\left|K\right|}{\left|H\right|}\implies\left|K\right|=\left[K:H\right]\left|H\right|$ .

Substituting the third equation into the first, we get

$\left|G\right|=\left[G:K\right]\left[K:H\right]\left|H\right|$ .

Now substituting this result into the second equation, we have

$\left[G:K\right]\left[K:H\right]\left|H\right|=\left[G:H\right]\left|H\right|\implies \left[G:H\right]=\left[G:K\right]\left[K:H\right]$ .

Does this make sense?

NonCommAlg · #4 $Old$ November 15th, 2009, 07:29 PM

a less trivial version of this problem is to let G be any group (not necessarily finite) and assume that [G : H] is finite. note that then both [G : K] and [K : H] will also be finite.